341Nishant sir, I dont know what the problem is. The solution is visible to me though.
number of possible ordered pair (s) (a,b) for each of which the equality a(cos x - 1) + b2 = cos(ax+b2) - 1 holds true for all x belonging to R are
1. 0
2. 1
3. 2
d. infinite
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9 Answers
62f(x)=cos x - 1
g(x) = ax+b^2
What i can see is f(g(x))=g(f(x))
Cant see anythign beyond this right now [2] except the feeling that the answer could probably be infinity!
cant think! (may be bcos of the fever :P)
341Since it is an identity, it should be true for x=0. On substituting this, we get b^2 = \cos b^2 -1 \le 0
This immediately gives us b = 0.
Hence we have for all x, a(\cos x -1) = \cos(ax) -1 \Rightarrow \cos (ax) - a \cos x + a-1=0
Differentiating, we get a (\sin (ax) - \sin x) =0
a=0 is a solution. Else we must have \sin (ax) - \sin x =0 for all x.
WLOG a>0. Now if we choose x such that 0<x<\frac{\pi}{2a}, then we have both x and ax lying in the interval \left(0, \frac{\pi}{2} \right)
For such x, \sin ax = \sin x \Rightarrow a=1 as sin x is strictly monotonic in that interval
Hence the only solutions are (0,0) and (1,0).
341
62visible is ok.. but did u understand it ;)
btw yeah i missed out on a sitter :P making things so much more complex :(